Differentiation

We have

velocity = ~ = 8* - $
.'. velocity when / = 5 is 40 — 5 = 35 ft. per second
To find the acceleration we have, if v be the velocity at time /,

acceleration = rate of increase of velocity with respect to time

dv
~7t
and since v = 8/ — 5

£-»

i.e. the acceleration is constant, and equal to 8 ft. per second per second.

100. EXAMPLE.—If ike pressure and volume of a gas are connectedly the relation

pv1 ;= C, where y and C are constants, find the volumetric elasticity eK = — v ~.

av
We have

PV* = C, and /. p = O

I = vy-

= 7.

= yp.

100a. EXAMPLE.—To test whether two quantities, y and x, are connected by «
relation of the form y = a + ^'"> having given a curve representing y as a function
of*.

Differentiating, we have

Taking logarithms

dx

log~~ = (// — i) log x + log bn.

Thus log j[~ and log x are connected by an equation of the first degree, and the

relation between them is represented by a straight line.
We therefore proceed as follows :—
From the curve tabulate values of y for equal intervals in the value of r.

Thence tabulate values of -£ as on p. 175.

ax

dy
Plot log -4- and log x. If the law is of the form y = a + bxn, we shall get a

straight line.

By substitution of the values of log ~f and log x obtained from two points on th*

line, the values of n — i and log bn are obtained, and n and b are determined.
The value of a may now be found by substitution in the equation y = a -f- b&

EXAMPLES.—LVIII.

Differentiate with respect to x.

1, 3;c2 — 2x 4- i. 2. 2 — $x — 6jc*.

3. 4^r6 — 3-#5 4 7^4 — •*'' + 2^2 — 4: — 3.

4. p:7 — 2i^4 4 6-t* — x. 6. 3JT3 — 2.r2 -f JT* — x®.